Question: Evaluate the double integral. $ \int_0^2 \int_1^{\sqrt{y}} x - 2y^2 \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{16}{3} - \dfrac{32\sqrt{2}}{7}$ (Choice B) B $\dfrac{8}{3} - \dfrac{16\sqrt{2}}{7}$ (Choice C) C $\dfrac{8}{3} - \dfrac{24\sqrt{2}}{7}$ (Choice D) D $\dfrac{16}{3} - \dfrac{24\sqrt{2}}{7}$
Explanation: First, we evaluate the inner integral. We can substitute in the $\sqrt{y}$ at the end as if it were a numerical bound. $\begin{aligned} & \int_0^2 \int_1^{\sqrt{y}} x - 2y^2 \, dx \, dy \\ \\ &= \int_0^2 \left[ \dfrac{x^2}{2} - 2xy^2 \right]_1^{\sqrt{y}} dy \\ \\ &= \int_0^2 \left( \dfrac{y}{2} - 2y^{2.5} \right) - \left( \dfrac{1}{2} - 2y^2 \right) \, dy \\ \\ &= \int_0^2 \dfrac{y}{2} - 2y^{2.5} - \dfrac{1}{2} + 2y^2 \, dy \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_0^2 \dfrac{y}{2} - 2y^{2.5} - \dfrac{1}{2} + 2y^2 \, dy &= \left[ \dfrac{y^2}{4} - \dfrac{2y^{3.5}}{3.5} - \dfrac{y}{2} + \dfrac{2y^3}{3} \right]_0^2 \\ \\ &= 1 - \dfrac{32 \sqrt{2}}{7} - 1 + \dfrac{16}{3} \\ \\ &= \dfrac{16}{3} - \dfrac{32 \sqrt{2}}{7} \end{aligned}$ The answer: $ \int_0^2 \int_1^{\sqrt{y}} x - 2y^2 \, dx \, dy = \dfrac{16}{3} - \dfrac{32 \sqrt{2}}{7}$